Problem: Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
using ull = unsigned long long;
using pii = pair<int, int>;
template <class T>
void cmax(T& x, const T& y) {
  if (x < y) x = y;
}
template <class T>
void cmin(T& x, const T& y) {
  if (x > y) x = y;
}
template <class T>
void sort(vector<T>& v) {
  sort(v.begin(), v.end());
}
template <class T>
void reverse(vector<T>& v) {
  reverse(v.begin(), v.end());
}
template <class T>
void unique(vector<T>& v) {
  sort(v.begin(), v.end()), v.erase(unique(v.begin(), v.end()), v.end());
}
void reverse(string& s) { reverse(s.begin(), s.end()); }
const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
  char ch;
  char getchar() {
    static char buf[io_size], *p1 = buf, *p2 = buf;
    return (p1 == p2) &&
                   (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  io_in& operator>>(char& c) {
    for (c = getchar(); isspace(c); c = getchar())
      ;
    return *this;
  }
  io_in& operator>>(string& s) {
    for (s.clear(); isspace(ch = getchar());)
      ;
    if (!~ch) return *this;
    for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
      ;
    return *this;
  }
  io_in& operator>>(char* str) {
    char* cur = str;
    while (*cur) *cur++ = 0;
    for (cur = str; isspace(ch = getchar());)
      ;
    if (!~ch) return *this;
    for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
      ;
    return *++cur = 0, *this;
  }
  template <class T>
  void read(T& x) {
    bool f = 0;
    while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);
    x = ~ch ? (ch ^ 48) : 0;
    while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
    x = f ? -x : x;
  }
  io_in& operator>>(int& x) { return read(x), *this; }
  io_in& operator>>(ll& x) { return read(x), *this; }
  io_in& operator>>(uint& x) { return read(x), *this; }
  io_in& operator>>(ull& x) { return read(x), *this; }
  io_in& operator>>(db& x) {
    read(x);
    bool f = x < 0;
    x = f ? -x : x;
    if (ch ^ '.') return *this;
    double d = 0.1;
    while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
    return x = f ? -x : x, *this;
  }
} in;
struct io_out {
  char buf[io_size], *s = buf;
  int pw[233], st[233];
  io_out() {
    set(7);
    for (register int i = (pw[0] = 1); i <= (9); ++i) pw[i] = pw[i - 1] * 10;
  }
  ~io_out() { flush(); }
  void io_chk() {
    if (s - buf > io_limit) flush();
  }
  void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }
  io_out& operator<<(char c) { return *s++ = c, *this; }
  io_out& operator<<(string str) {
    for (char c : str) *s++ = c;
    return io_chk(), *this;
  }
  io_out& operator<<(char* str) {
    char* cur = str;
    while (*cur) *s++ = *cur++;
    return io_chk(), *this;
  }
  template <class T>
  void write(T x) {
    if (x < 0) *s++ = '-', x = -x;
    do {
      st[++st[0]] = x % 10, x /= 10;
    } while (x);
    while (st[0]) *s++ = st[st[0]--] ^ 48;
  }
  io_out& operator<<(int x) { return write(x), io_chk(), *this; }
  io_out& operator<<(ll x) { return write(x), io_chk(), *this; }
  io_out& operator<<(uint x) { return write(x), io_chk(), *this; }
  io_out& operator<<(ull x) { return write(x), io_chk(), *this; }
  int len, lft, rig;
  void set(int _length) { len = _length; }
  io_out& operator<<(db x) {
    bool f = x < 0;
    x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
    return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
  }
} out;
template <long long sz, long long mod>
struct math_t {
  math_t() {
    fac.resize(sz + 1), ifac.resize(sz + 1);
    for (register long long i = (fac[0] = 1); i <= (sz); ++i)
      fac[i] = fac[i - 1] * i % mod;
    ifac[sz] = inv(fac[sz]);
    for (register long long i = (sz - 1); i >= (0); --i)
      ifac[i] = ifac[i + 1] * (i + 1) % mod;
  }
  vector<long long> fac, ifac;
  long long qpow(long long x, long long y) {
    long long ans = 1;
    for (; y; y >>= 1, x = x * x % mod)
      if (y & 1) ans = ans * x % mod;
    return ans;
  }
  long long inv(long long x) { return qpow(x, mod - 2); }
  long long C(long long n, long long m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
  }
};
long long gcd(long long x, long long y) { return !y ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long h, g;
const long long maxn = 1 << 22;
long long a[maxn], b[maxn], dep[maxn];
long long qwq[maxn];
long long quq = 0, cnt = 0;
long long st[maxn], top = 0;
void Func(long long p) {
  a[st[++top] = p] = 0;
  if (a[p << 1] || a[p << 1 | 1]) {
    if (a[p << 1] > a[p << 1 | 1])
      a[p] = a[p << 1], Func(p << 1);
    else
      a[p] = a[p << 1 | 1], Func(p << 1 | 1);
  }
  if (!a[p] && dep[p] <= g) quq = 1;
}
void put(long long p) {
  if (quq) {
    while (top) a[st[top]] = b[st[top]], top--;
  } else {
    while (top) b[st[top]] = a[st[top]], top--;
    qwq[++cnt] = p;
  }
}
void dfs(long long p, long long n) {
  if (p < n) {
    while (a[p] && !quq) {
      Func(p);
      put(p);
    }
    quq = 0;
    dfs(p << 1, n);
    dfs(p << 1 | 1, n);
  } else
    return;
}
signed main() {
  for (long long i = 1; i < maxn; i++) dep[i] = dep[i >> 1] + 1;
  long long _;
  in >> _;
  auto solve = [&](long long n) {
    dfs(1, n);
    long long m = 1 << g;
    m--;
    long long ans = 0;
    for (register long long i = (1); i <= (m); ++i) ans += a[i];
    out << ans << '\n';
    for (register long long i = (1); i <= (cnt); ++i) out << qwq[i] << ' ';
    out << '\n';
    cnt = 0;
    for (register long long i = (1); i <= (n); ++i) a[i] = b[i] = 0;
  };
  while (_--) {
    in >> h >> g;
    long long n = 1 << h;
    n--;
    for (register long long i = (1); i <= (n); ++i) in >> a[i];
    for (register long long i = (1); i <= (n); ++i) b[i] = a[i];
    solve(n);
  }
  return 0;
}